If y = \(x+\frac{1}{x}\) then x4 + x3 – 4x2 + x + 1 = 0 can be reduc

1.	If y = \(x+\frac{1}{x}\) then x4 + x3 – 4x2 + x + 1 = 0 can be reduced to which one of the following ? (x ≠ 0)(a) y2 + y – 2 = 0 (b) y2 + y – 4 = 0 (c) y2 + y – 6 = 0 (d) y2 + y + 6 = 0
Answer» (c) y² + y – 6 = 0 x⁴ + x³ – 4x² + x + 1 = 0 ⇒ x² + x - 4 + \(\frac{1}{x}\) + \(\frac{1}{x^2}\) = 0 ⇒ \(x^2+\frac{1}{x^2}\) + \(x+\frac{1}{x}\) - 4 = 0 ⇒ \(\big(x+\frac{1}{x}\big)^2\) - 2 + \(\big(x+\frac{1}{x}\big)\) - 4 = 0 ⇒ y² + y – 6 = 0 \(\big(∵x+\frac{1}{x}=y\big)\)

If y = \(x+\frac{1}{x}\) then x4 + x3 – 4x2 + x + 1 = 0 can be reduced to which one of the following ? (x ≠ 0)(a) y2 + y – 2 = 0 (b) y2 + y – 4 = 0 (c) y2 + y – 6 = 0 (d) y2 + y + 6 = 0

Answer»

(c) y² + y – 6 = 0

x⁴ + x³ – 4x² + x + 1 = 0

⇒ x² + x - 4 + \(\frac{1}{x}\) + \(\frac{1}{x^2}\) = 0

⇒ \(x^2+\frac{1}{x^2}\) + \(x+\frac{1}{x}\) - 4 = 0

⇒ \(\big(x+\frac{1}{x}\big)^2\) - 2 + \(\big(x+\frac{1}{x}\big)\) - 4 = 0

⇒ y² + y – 6 = 0 \(\big(∵x+\frac{1}{x}=y\big)\)